Question: Define $f(x, y) = x^2 - y^2$. Let $\vec{a} = (-1, 1)$ and $\vec{v} = \left( \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Our ultimate goal is to substitute $h = 0$ into the limit and get the answer. To do this, we first need to cancel out the $h$ in the denominator. Otherwise we're dividing by zero. How can we simplify the limit so that we get an $h$ on the top and an $h$ on the bottom, which we can then cancel? Let's plug in $\vec{a}$ and $\vec{v}$ to the limit. $ \lim_{h \to 0} \dfrac{f \left( (-1, 1) + h \left( \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right) \right) - f(-1, 1)}{h}$ Now we can add together the vectors. $ \lim_{h \to 0} \dfrac{f \left(-1 + \dfrac{\sqrt{2}}{2} h, 1 + \dfrac{\sqrt{2}}{2} h \right) - f(-1, 1)}{h}$ Let's evaluate $f$. $ \lim_{h \to 0} \dfrac{\left( -1 + \dfrac{\sqrt{2}}{2} h \right)^2 - \left( 1 + \dfrac{\sqrt{2}}{2} h \right)^2 - (1 - 1)}{h}$ When we simplify, we can cancel out all the terms without an $h$. $ \lim_{h \to 0} \dfrac{1 - h \sqrt{2} + \dfrac{2}{4}h^2 - 1 - h\sqrt{2} + \dfrac{\sqrt{2}}{4}h^2}{h}$ becomes $ \lim_{h \to 0} \dfrac{(-2\sqrt{2} + h)h}{h}$ Now we can cancel $h$ and calculate the limit. $\begin{aligned} \lim_{h \to 0} \dfrac{(-2\sqrt{2} + h)h}{h} &= \lim_{h \to 0} -2\sqrt{2} + h \\ \\ &= -2\sqrt{2} \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = -2\sqrt{2}$.